Difference between revisions of "Symmedians, Lemoine point"
(→Symmedian and tangents) |
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'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
==Symmedian and tangents== | ==Symmedian and tangents== | ||
− | [[File:Tangents and symmedian.png| | + | [[File:Tangents and symmedian.png|220px|right]] |
− | Let <math>\triangle ABC</math> and it’s circumcircle <math>\Omega</math> be given. Tangents to <math>\Omega</math> at points <math>B</math> and <math>C</math> intersect at point <math>F.</math> | + | Let <math>\triangle ABC</math> and it’s circumcircle <math>\Omega</math> be given. |
+ | |||
+ | Tangents to <math>\Omega</math> at points <math>B</math> and <math>C</math> intersect at point <math>F.</math> | ||
+ | |||
Prove that <math>AF</math> is <math>A-</math> symmedian of <math>\triangle ABC.</math> | Prove that <math>AF</math> is <math>A-</math> symmedian of <math>\triangle ABC.</math> | ||
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<cmath>\triangle FDC \sim \triangle FCA \implies \frac {CD}{AC} = \frac{DF}{CF}.</cmath> | <cmath>\triangle FDC \sim \triangle FCA \implies \frac {CD}{AC} = \frac{DF}{CF}.</cmath> | ||
<math>BF = CF \implies \frac {BD}{CD} = \frac{AB}{AC} \implies AD</math> is <math>A-</math> symmedian of <math>\triangle ABC.</math> | <math>BF = CF \implies \frac {BD}{CD} = \frac{AB}{AC} \implies AD</math> is <math>A-</math> symmedian of <math>\triangle ABC.</math> | ||
+ | |||
+ | <i><b>Corollary</b></i> | ||
+ | |||
+ | [[File:Tangents to symmedian.png|220px|right]] | ||
+ | Let <math>\triangle ABC</math> and it’s circumcircle <math>\Omega</math> be given. | ||
+ | |||
+ | Let tangent to <math>\Omega</math> at points <math>A</math> intersect line <math>BC</math> at point <math>F.</math> | ||
+ | |||
+ | Let <math>FD</math> be the tangent to <math>\Omega</math> different from <math>FA.</math> | ||
+ | |||
+ | Then <math>AD</math> is <math>A-</math> symmedian of <math>\triangle ABC.</math> | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Revision as of 13:55, 14 July 2024
The reflecting of the median over the corresponding angle bisector is the symmedian. The angle formed by the symmedian and the angle bisector has the same measure as the angle between the median and the angle bisector, but it is on the other side of the angle bisector. The symmedian is isogonally conjugate to the median
There are three symmedians. They are meet at a triangle center called the Lemoine point.
Proportions
Let be given.
Let be the median,
Prove that iff is the symmedian than
Proof
1. Let be the symmedian. So
Similarly
By applying the Law of Sines we get
Similarly,
2.
As point moves along the fixed arc
from
to
, the function
monotonically increases from zero to infinity. This means that there is exactly one point at which the condition is satisfied. In this case, point
lies on the symmedian.
Similarly for point
Corollary
Let be the
symmedian of
Then is the
symmedian of
is the
symmedian of
is the
symmedian of
vladimir.shelomovskii@gmail.com, vvsss
Symmedian and tangents
Let and it’s circumcircle
be given.
Tangents to at points
and
intersect at point
Prove that is
symmedian of
Proof
Denote WLOG,
is
symmedian of
Corollary
Let and it’s circumcircle
be given.
Let tangent to at points
intersect line
at point
Let be the tangent to
different from
Then is
symmedian of
vladimir.shelomovskii@gmail.com, vvsss